# What is the maximum number of points of intersection of 8 circles?

I came across this question on a book on combinations and permutations. Although it had a vaguely defined one step answer : 8P2 = 56. As far as I could understand, the above step gives the number of arrangements possible for 8 objects taken two at a time. But I don't understand how it gives the points of intersection. Also, is there any other method to solve this kind of problems ? Regards

• This is rather confusing. Are all the circles intersecting at the same points, or just any circles intersecting? I agree that the latter case seems unintuitive if the answer is 28. – theREALyumdub Nov 13 '15 at 17:16
• @theREALyumdub Can you please explain, considering latter is the case. Thanks – H G Sur Nov 13 '15 at 17:17
• Isn't 8P2 $56$? – user2357112 supports Monica Nov 13 '15 at 17:47
• For the pedantic, you might want to restrict the problem so that none of the circles are the same. – Rick Decker Nov 13 '15 at 21:32

Two circles can intersect in at most two points (excluding the case where two circles coincide with each other). So by looking at all possible pairs of circles, the maximum number of intersection points is given by $2\binom{8}{2}=56$. (A simple proof of the need for the multiplier of two is that two circles can intersect in 2 points, whereas $\binom{2}{2}=1$).

Now consider the case where all circles are the same size and close enough together that the center of each circle lies inside every other circle. In the general case where none of the pairwise intersection points lies on the perimeter of any of the six other circles, the maximum number of intersection points is realisable.

Two circles can intersect in at most two points. So if you have $8$ circles for every pair there can be at most $2$ intersections giving us the answer $2\cdot{8\choose 2} = 56$.

However this does in and of itself not show that this number can also be reached. There might after all are points where $3$ circles meet lowering the total number of intersections.

Fortunately it is easy to construct $8$ circles that have no triple intersections. For example you can place $8$ circles all with radius $8$ on the coordinates $(0,0), (1,0),\ldots,(7,0)$. All these circles clearly intersect twice with each other. But if there was some point $x$ where three circles intersected then the circle with radius $8$ around $x$ would intersect the $x$-axis in $3$ points, which is clearly absurd.

There's a simple answer that probably anyone could understand.

Let's take the example of eight 10" circles placed in a line along the X-axis. Let's arrange these so that their center is along the X-axis, and their centers are at (0,0), (1,0), (2,0), (3,0)...(7,0). That is they all overlap and cross in two points with all the other circles.

So, if each of the 8 circles crosses the other 7 circles twice, that would be 7x8x2. But wait! When this is done, you end up counting each pair of circles twice, ie. Circle 1 crosses circle 5 and circle 5 crosses circle 1. So the answer is really 7x8 or 56.